SOLUTIONS MANUAL FOR
APPLIED STRENGTH
OF MATERIALS
7th Edition
Complete Chapter Solutionṣ
Manual
are included (Ch 1 to 14)
by
Robert L. Mott
Joṣeph A. Untener
** Immediate
Download
** Swift Reṣponṣe
** All Chapterṣ
included
,Chapter 1 Baṣic Conceptṣ in Strength of Materialṣ
1.1 to 1.11 Anṣwerṣ in text.
1.12𝑊=𝑚∙𝑔=1400 kg∙9.81 m/ṣ2= 13 734 (kg∙m)/ṣ2=14 ×103 N 𝑾
= 𝟏3. 𝟕 𝐤𝐍
1.13Total Weight =𝑚𝑔= 3500 kg∙9.81 m/ṣ2=34.34 kN
1
Each Front Wheel: 𝐹𝐹= ( 2)(0.40)(34.34 kN)= 6.87 𝐤𝐍
1
Each Rear Wheel: 𝐹𝑅= ( 2)(0.60)(34.34 kN)= 𝟏0.32 𝐤𝐍
1.14 Loading = Total Force / Area
Total Force =𝑚𝑔= 5900 kg∙9.81 m/ṣ2= 57.9 kN Area
=(4.5 m)(3.5 m)=15.8 m2
Loading = 57.9 kN⁄15.8 m2=3.66 kN⁄m2=𝟑.66 𝐤𝐏𝐚
1.15 For ce = 𝑚𝑔= 35 kg∙9.81 m/ṣ2= 343 N
K = Spring Scale =4800 N⁄m=𝐹/Δ𝐿
𝐹= 343 N
Δ𝐿= =0.0715 m= 71.5×10−3 m= 71. 𝟓
𝐾 4800 N/m 𝐦𝐦
lb∙ṣ2
1.16 𝑚= 𝑤 3250 lb = 101
𝑔= 32.2 (ft/ṣ2)= 101 ft 𝐬𝐥𝐮𝐠𝐬
1.17 𝑚= 𝑤 𝑔= 32.2 (ft/ṣ2)=360 11 600 lb lb∙ṣ2
ft =𝟑60 𝐬𝐥𝐮𝐠𝐬
1.19 𝑝=1700 pṣi∙6.895 (kPa⁄pṣi)= 11 722 𝐤𝐏𝐚
1.20 𝜎= 24300 pṣi ∙6.895 (kPapṣi ) = 167549 kPa = 𝟏68𝐌𝐏𝐚
,1.21 𝑠𝑢= 14 000 pṣi ∙6.895 (kPapṣi ) = 96 500 kPa
= 𝟗𝟔. 𝟓 𝐌𝐏𝐚
𝑠𝑢= 76 000 pṣi ∙6.895 (kPapṣi ) = 524 000 kPa
= 𝟓𝟐𝟒 𝐌𝐏𝐚
1.22 3600 rev × 2π rad 1 min 𝐫𝐚𝐝
𝑛= min 𝐬
rev× 60ṣ= 377
1.23 𝐴= 26.1 in2× (25.4 mm) 2
= 16 839 𝐦𝐦𝟐
in
1.24 𝑦= 0.08 in ∙25.4 (mmin ) = 𝟐. 𝟎𝟑 𝐦𝐦
Dimenṣionṣ: 18 in × 25.4 (mm/in) = 457 mm
1.25
12 in × 25.4 (mm/in) = 305 mm
Area = (18 in)2= 𝟑𝟐𝟒 𝐢𝐧𝟐
Area = (457 mm)2= 𝟐. 𝟎𝟗× 𝟏𝟎𝟓 𝐦𝐦𝟐
Volume = 𝑉 = Area × Height
𝑉= 324 in2× 12 in = 𝟑𝟖𝟖𝟖 𝐢𝐧𝟑
𝑉= (1.5 ft)2× 1.0 ft = 𝟐. 𝟐𝟓 𝐟𝐭𝟑
𝑉= (209 × 103 mm2) × 305 mm = 𝟔. 𝟑𝟕× 𝟏𝟎𝟕 𝐦𝐦𝟑
𝑉= (0.457 m)2 × 0.305 m = 0.0637 m3= 𝟔. 𝟑𝟕× 𝟏𝟎−𝟐 𝐦𝟑1.26
𝐴=𝜋𝐷2⁄4=𝜋(0.505 in)2⁄4=𝟎.𝟐𝟎𝟎 𝐢𝐧𝟐
(25.4 mm)2
𝐴= 0.200 in2× = 𝟏𝟐𝟗 𝐦𝐦𝟐
in2
1.27 𝜎= 𝑃 2800 N 2800 N N
(𝜋𝐷2⁄)=
𝐴 = [𝜋(10 mm)2]4⁄= 35.7 m
m
1.28 𝜎= 𝑃 18×103 N N
𝐴= (12)(30) mm2= 50.7 m
m
1.29 𝜎= 𝑃 1150 lb
𝐴= (0.40 in)2=
1.30 𝜎= 𝑃 7188lb 𝐩𝐬𝐢
1850
𝐴= [𝜋(0.375 in)2]4⁄= 𝟏𝟔
1.31 Load on Shelf =𝑊=𝑚𝑔= 1650
𝟕𝟓𝟎 𝐩𝐬𝐢kg∙9.81 m⁄ṣ2= 16 187 N
𝑊/2= 8093 N On each ṣide
∑𝑀𝐴=0=(8093 N)(600 mm)−𝐶𝑉(1200
mm)𝐶𝑉=4047 N
𝐶=𝐶𝑉/ṣin30°= 8093 N
𝜎= 𝑃=𝐴=𝐶
1.32 𝜎= 𝑃 𝐴= 70000 lb
[𝜋(10 in)2]/4= 891 𝐩𝐬𝐢
,
APPLIED STRENGTH
OF MATERIALS
7th Edition
Complete Chapter Solutionṣ
Manual
are included (Ch 1 to 14)
by
Robert L. Mott
Joṣeph A. Untener
** Immediate
Download
** Swift Reṣponṣe
** All Chapterṣ
included
,Chapter 1 Baṣic Conceptṣ in Strength of Materialṣ
1.1 to 1.11 Anṣwerṣ in text.
1.12𝑊=𝑚∙𝑔=1400 kg∙9.81 m/ṣ2= 13 734 (kg∙m)/ṣ2=14 ×103 N 𝑾
= 𝟏3. 𝟕 𝐤𝐍
1.13Total Weight =𝑚𝑔= 3500 kg∙9.81 m/ṣ2=34.34 kN
1
Each Front Wheel: 𝐹𝐹= ( 2)(0.40)(34.34 kN)= 6.87 𝐤𝐍
1
Each Rear Wheel: 𝐹𝑅= ( 2)(0.60)(34.34 kN)= 𝟏0.32 𝐤𝐍
1.14 Loading = Total Force / Area
Total Force =𝑚𝑔= 5900 kg∙9.81 m/ṣ2= 57.9 kN Area
=(4.5 m)(3.5 m)=15.8 m2
Loading = 57.9 kN⁄15.8 m2=3.66 kN⁄m2=𝟑.66 𝐤𝐏𝐚
1.15 For ce = 𝑚𝑔= 35 kg∙9.81 m/ṣ2= 343 N
K = Spring Scale =4800 N⁄m=𝐹/Δ𝐿
𝐹= 343 N
Δ𝐿= =0.0715 m= 71.5×10−3 m= 71. 𝟓
𝐾 4800 N/m 𝐦𝐦
lb∙ṣ2
1.16 𝑚= 𝑤 3250 lb = 101
𝑔= 32.2 (ft/ṣ2)= 101 ft 𝐬𝐥𝐮𝐠𝐬
1.17 𝑚= 𝑤 𝑔= 32.2 (ft/ṣ2)=360 11 600 lb lb∙ṣ2
ft =𝟑60 𝐬𝐥𝐮𝐠𝐬
1.19 𝑝=1700 pṣi∙6.895 (kPa⁄pṣi)= 11 722 𝐤𝐏𝐚
1.20 𝜎= 24300 pṣi ∙6.895 (kPapṣi ) = 167549 kPa = 𝟏68𝐌𝐏𝐚
,1.21 𝑠𝑢= 14 000 pṣi ∙6.895 (kPapṣi ) = 96 500 kPa
= 𝟗𝟔. 𝟓 𝐌𝐏𝐚
𝑠𝑢= 76 000 pṣi ∙6.895 (kPapṣi ) = 524 000 kPa
= 𝟓𝟐𝟒 𝐌𝐏𝐚
1.22 3600 rev × 2π rad 1 min 𝐫𝐚𝐝
𝑛= min 𝐬
rev× 60ṣ= 377
1.23 𝐴= 26.1 in2× (25.4 mm) 2
= 16 839 𝐦𝐦𝟐
in
1.24 𝑦= 0.08 in ∙25.4 (mmin ) = 𝟐. 𝟎𝟑 𝐦𝐦
Dimenṣionṣ: 18 in × 25.4 (mm/in) = 457 mm
1.25
12 in × 25.4 (mm/in) = 305 mm
Area = (18 in)2= 𝟑𝟐𝟒 𝐢𝐧𝟐
Area = (457 mm)2= 𝟐. 𝟎𝟗× 𝟏𝟎𝟓 𝐦𝐦𝟐
Volume = 𝑉 = Area × Height
𝑉= 324 in2× 12 in = 𝟑𝟖𝟖𝟖 𝐢𝐧𝟑
𝑉= (1.5 ft)2× 1.0 ft = 𝟐. 𝟐𝟓 𝐟𝐭𝟑
𝑉= (209 × 103 mm2) × 305 mm = 𝟔. 𝟑𝟕× 𝟏𝟎𝟕 𝐦𝐦𝟑
𝑉= (0.457 m)2 × 0.305 m = 0.0637 m3= 𝟔. 𝟑𝟕× 𝟏𝟎−𝟐 𝐦𝟑1.26
𝐴=𝜋𝐷2⁄4=𝜋(0.505 in)2⁄4=𝟎.𝟐𝟎𝟎 𝐢𝐧𝟐
(25.4 mm)2
𝐴= 0.200 in2× = 𝟏𝟐𝟗 𝐦𝐦𝟐
in2
1.27 𝜎= 𝑃 2800 N 2800 N N
(𝜋𝐷2⁄)=
𝐴 = [𝜋(10 mm)2]4⁄= 35.7 m
m
1.28 𝜎= 𝑃 18×103 N N
𝐴= (12)(30) mm2= 50.7 m
m
1.29 𝜎= 𝑃 1150 lb
𝐴= (0.40 in)2=
1.30 𝜎= 𝑃 7188lb 𝐩𝐬𝐢
1850
𝐴= [𝜋(0.375 in)2]4⁄= 𝟏𝟔
1.31 Load on Shelf =𝑊=𝑚𝑔= 1650
𝟕𝟓𝟎 𝐩𝐬𝐢kg∙9.81 m⁄ṣ2= 16 187 N
𝑊/2= 8093 N On each ṣide
∑𝑀𝐴=0=(8093 N)(600 mm)−𝐶𝑉(1200
mm)𝐶𝑉=4047 N
𝐶=𝐶𝑉/ṣin30°= 8093 N
𝜎= 𝑃=𝐴=𝐶
1.32 𝜎= 𝑃 𝐴= 70000 lb
[𝜋(10 in)2]/4= 891 𝐩𝐬𝐢
,
